# [seqfan] A new and stubborn "Yellowstone" sequence

Neil Sloane njasloane at gmail.com
Fri Mar 3 02:45:39 CET 2017

```Dear Seq Fans:
Some three years ago the Yellowstone permutation A098550 was discussed
on this list. Others of the same type are the EKG sequence A064413,
Quet's A127202, and Sigrist's A280985. In all 4 cases the sequence can
be shown to be a permutation of the natural numbers. Another example
which appears to be a permutation is A055265, although that one seems
much too difficult.

I've been spending too much time lately on a new one, also from
Remy Sigrist, A280864: the lexicographically earliest sequence with
the property that if a prime p divides a(n) then p also divides
exactly one of a(n-1) and a(n+1). It starts (beginning with a(1)):
1, 2, 4, 3, 6, 8, 5, 10, 12, 9, 7, 14, 16, 11, 22, 18, 15, ...
I can "almost" prove it is a permutation! Here is what I can prove
(see A280864 for details):
- every prime appears (naked)
- every even number appears
- there are infinitely many odd composite terms
but I cannot prove that every odd number appears
-------------------------
Today I wrote out the full proof that A127202 is a permutation,
which is attached as a .txt file to that entry.
I was hoping it would help with A280864, but ...
Read on only if you are interested in that sequence.
-------------------------
I have dozens of pages of notes on A280864, but I'll just mention a few things:
Call a(n) "satisfied" if all its prime factors already divide a(n-1).
If a(n) is satisfied, a(n+1) is the smallest missing number relatively
prime to a(n).
It is helpful to split the primes p into two classes:
Type I. The first time that p divides any term is after a satisfied
term a(n-1), with a(n) = p, a(n+1) = 2p, a(n+2) = smallest missing
even number.
Type II. The first time that p divides any term is when a(n) = b*p,
b>1, a(n+1) = p (and is satisfied).
The Type II primes (A280745) are
13, 139, 379, 397, 647, 661, 967, 983, 997, 1021, 1063, 1109,...
and there appear to be infinitely many of them. It appears that most
primes are Type I, and I can prove there are infinitely many of Type
I.
If we knew there were infinitely many of Type II we would be done,
because after a Type II prime p we pick up the smallest missing number
relatively prime to p, and p keeps increasing.

Because every even number appears, every 2^r appears, and if the term
before 2^r was always even (as seems to be the case, see A282024), 2^r
would be satisfied and if that happened infinitely often we would pick
up all the missing odd numbers at the following terms and we would be
done.
The sequence giving the smallest missing terms is A280740, and seems
to consist of primes except that 25 appears four times.

There are at least 19 related sequences in the OEIS now, also an
"a-file" for A280864 with a million terms.
-------------------------
My feeling is that the way to finish the proof - that is, to prove
that every odd number appears - is to focus on the 2^r terms. We can
assume there are infinitely many terms 2^r which are immediately
preceded by an odd number (d_r say) (or else we would be done).
Suppose m is a missing odd composite number, and go out a long way in
the sequence until all the terms are huge, and look at a(n-2),
a(n-1)=d_r, a(n)=2^r.
Then we could have used m instead of 2^r (getting a contradiction)
unless some prime factor q of m divides both a(n-2) and a(n-1). And
this must be true for every term 2^r from some point on, and the same
thing must be true for every missing odd number. I feel that this is
only a few cups of coffee away from completing the proof, but ...
```